Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $n = \dfrac{6y + 48}{y - 2} \div \dfrac{y^2 + y - 56}{y^2 - 2y} $
Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{6y + 48}{y - 2} \times \dfrac{y^2 - 2y}{y^2 + y - 56} $ First factor the quadratic. $n = \dfrac{6y + 48}{y - 2} \times \dfrac{y^2 - 2y}{(y + 8)(y - 7)} $ Then factor out any other terms. $n = \dfrac{6(y + 8)}{y - 2} \times \dfrac{y(y - 2)}{(y + 8)(y - 7)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ 6(y + 8) \times y(y - 2) } { (y - 2) \times (y + 8)(y - 7) } $ $n = \dfrac{ 6y(y + 8)(y - 2)}{ (y - 2)(y + 8)(y - 7)} $ Notice that $(y - 2)$ and $(y + 8)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ 6y\cancel{(y + 8)}(y - 2)}{ (y - 2)\cancel{(y + 8)}(y - 7)} $ We are dividing by $y + 8$ , so $y + 8 \neq 0$ Therefore, $y \neq -8$ $n = \dfrac{ 6y\cancel{(y + 8)}\cancel{(y - 2)}}{ \cancel{(y - 2)}\cancel{(y + 8)}(y - 7)} $ We are dividing by $y - 2$ , so $y - 2 \neq 0$ Therefore, $y \neq 2$ $n = \dfrac{6y}{y - 7} ; \space y \neq -8 ; \space y \neq 2 $